Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $t = \dfrac{y + 5}{y^2 + 12y + 35} \div \dfrac{y - 6}{y^2 - 49} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $t = \dfrac{y + 5}{y^2 + 12y + 35} \times \dfrac{y^2 - 49}{y - 6} $ First factor out any common factors. $t = \dfrac{y + 5}{y^2 + 12y + 35} \times \dfrac{y^2 - 49}{y - 6} $ Then factor the quadratic expressions. $t = \dfrac {y + 5} {(y + 7)(y + 5)} \times \dfrac {(y + 7)(y - 7)} {y - 6} $ Then multiply the two numerators and multiply the two denominators. $t = \dfrac {(y + 5) \times (y + 7)(y - 7) } { (y + 7)(y + 5) \times (y - 6)} $ $t = \dfrac {(y + 7)(y - 7)(y + 5)} {(y + 7)(y + 5)(y - 6)} $ Notice that $(y + 7)$ and $(y + 5)$ appear in both the numerator and denominator so we can cancel them. $t = \dfrac {\cancel{(y + 7)}(y - 7)(y + 5)} {\cancel{(y + 7)}(y + 5)(y - 6)} $ We are dividing by $y + 7$ , so $y + 7 \neq 0$ Therefore, $y \neq -7$ $t = \dfrac {\cancel{(y + 7)}(y - 7)\cancel{(y + 5)}} {\cancel{(y + 7)}\cancel{(y + 5)}(y - 6)} $ We are dividing by $y + 5$ , so $y + 5 \neq 0$ Therefore, $y \neq -5$ $t = \dfrac {y - 7} {y - 6} $ $ t = \dfrac{y - 7}{y - 6}; y \neq -7; y \neq -5 $